从 dataframe 列值的平均值和标准偏差创建表 R

Question

我是 R 的新手，所以不知道代码。 我有两个数据框。 一个 dataframe 看起来像这样。

df

ID	疾病
GSM239170	控制
GSM239323	控制
GSM239324	控制
GSM239326	控制
GSM239328	反洗钱
GSM239329	反洗钱
GSM239331	反洗钱
GSM239332	控制
GSM239333	控制

而另一个 dataframe 看起来像这样：

df1

GSM239170	GSM239323	GSM239324	GSM239326	GSM239328	GSM239329	GSM239331	GSM239332	GSM239333
3.016704177	3.285669072	2.929482692	2.922820483	3.15950317	3.163327169	2.985901308	3.122708843	3.070948463
7.977735461	6.532514237	6.388007183	6.466679556	6.432795021	6.407321524	6.426470803	6.376394357	6.469070308
4.207280707	4.994965767	4.40159671	4.747114589	4.830045513	4.213762092	4.884418365	4.4318876	4.849665444
7.25609471	7.420807337	6.999340125	7.094488581	7.024332721	7.17928981	7.159898654	7.009977785	6.830979234
2.204955099	2.331625217	2.133305231	2.18332885	2.12778313	2.269697813	2.264705552	2.253940441	2.287924323
7.28437278	6.983593721	6.86337111	6.865970678	7.219840938	7.181113053	7.392230178	7.484052914	7.52498281
4.265792764	4.970684112	4.595545125	4.575545289	4.547957809	4.68215122	4.674495889	4.675841709	4.643311767
2.6943516	2.916324936	2.578130269	2.659717988	2.567436676	2.8095128	2.790110381	2.795882913	2.884588792
3.646303109	8.817891552	11.4248793	10.74738082	9.296043108	9.53150669	8.285160496	9.769919327	9.774610531
3.040292001	3.38486713	2.958851115	3.047880699	2.878562717	3.209319974	3.20260379	3.195993624	3.3004227
2.357625231	2.444753172	2.340767158	2.32143889	2.282608342	2.401218719	2.385568421	2.375334953	2.432634747
5.378494673	6.065038394	5.134842087	5.367342376	5.682051149	5.712072512	5.57179966	5.72082395	5.656674512
2.833814735	3.038434511	2.837711812	2.859800224	2.866040813	2.969167906	2.929449968	2.963530689	2.931065261
6.192932281	6.478439634	6.180169144	6.151689376	6.238949956	6.708196123	6.441437631	6.448280595	6.413562269
4.543042482	4.786227217	4.445131477	4.51471011	4.491645167	4.460114204	4.602482637	4.587221948	4.623125028
6.069437462	6.232738284	6.74644117	7.04995802	6.938928532	6.348253102	6.080950712	6.324619355	6.472893789

我想制作一个表格，包括每个基因的 mean_AML、sd_AML（标准差）、min_AML、max_AML、mean_Control、sd_Control、min_Control、max_Control 和 Fold_change（即 mean_AML – mean_Control）。 使用内置函数很好。

无法弄清楚我该怎么做。 请帮忙。

谢谢。

Answer 1

我们可以将pivot_longer与right_join结合起来，然后在组上使用summarise ：

library(dplyr)
library(tidyr)
df1 %>% 
  pivot_longer(
    everything(),
    names_to = "ID", 
    values_to = "value"
  ) %>% 
  right_join(df, by="ID") %>% 
  group_by(Disease) %>% 
  summarise(Min = min(value), Mean = mean(value), Max = max(value), Sd = sd(value)) %>%
  ungroup()

  Disease   Min  Mean   Max    Sd
  <chr>   <dbl> <dbl> <dbl> <dbl>
1 AML      2.13  4.91  9.53  2.04
2 Control  2.13  4.92 11.4   2.12

Answer 2

使用pivot_longer()将第二个 dataframe 放在长格式中会有所帮助：

library(tidyr)
newdf <- pivot_longer(df1, 
    cols = everything(),
    names_to = 'ID',
    values_to = 'value')

然后将两个数据框merge()为一个：

df.all <- merge(df, newdf, by = 'ID', all = T)

然后你可以group_by(gene)和summarise() ，添加你需要的任何值：

library(dplyr)
df.all %>% group_by(gene) %>% summarise(sd = sd(value), min = min(value), max = max(value), mean = mean(value))

Answer 3

旧的 function tidyr::gather的另一个选项有一个列 df：

library(tidyverse)

df2_spread <- df1 %>% 
  tidyr::gather(ID, val) %>% 
  left_join(df, by = 'ID')
  
df2_spread %>% 
  group_by(Disease, gene = ID) %>% 
  summarise(n = n(),
            mean = mean(val),
            sd = sd(val),
            min = min(val),
            max = max(val), .groups = "drop")

 A tibble: 9 × 7
  Disease gene          n  mean    sd   min   max
  <chr>   <chr>     <int> <dbl> <dbl> <dbl> <dbl>
1 AML     GSM239328    16  4.91  2.15  2.13  9.30
2 AML     GSM239329    16  4.95  2.13  2.27  9.53
3 AML     GSM239331    16  4.88  1.96  2.26  8.29
4 Control GSM239170    16  4.56  1.91  2.20  7.98
5 Control GSM239323    16  5.04  1.98  2.33  8.82
6 Control GSM239324    16  4.93  2.45  2.13 11.4 
7 Control GSM239326    16  4.97  2.34  2.18 10.7 
8 Control GSM239332    16  4.97  2.16  2.25  9.77
9 Control GSM239333    16  5.01  2.14  2.29  9.77

在任何情况下，我都无法找到一种方法来计算每个基因的 Fold_change，因为这里似乎只有一种基因疾病。

这是数据

df <- tibble::tribble(
          ~ID,  ~Disease,
  "GSM239170", "Control",
  "GSM239323", "Control",
  "GSM239324", "Control",
  "GSM239326", "Control",
  "GSM239328",     "AML",
  "GSM239329",     "AML",
  "GSM239331",     "AML",
  "GSM239332", "Control",
  "GSM239333", "Control"
  )



df1 <- tibble::tribble(
   ~GSM239170,  ~GSM239323,  ~GSM239324,  ~GSM239326,  ~GSM239328,  ~GSM239329,  ~GSM239331,  ~GSM239332,  ~GSM239333,
  3.016704177, 3.285669072, 2.929482692, 2.922820483,  3.15950317, 3.163327169, 2.985901308, 3.122708843, 3.070948463,
  7.977735461, 6.532514237, 6.388007183, 6.466679556, 6.432795021, 6.407321524, 6.426470803, 6.376394357, 6.469070308,
  4.207280707, 4.994965767,  4.40159671, 4.747114589, 4.830045513, 4.213762092, 4.884418365,   4.4318876, 4.849665444,
   7.25609471, 7.420807337, 6.999340125, 7.094488581, 7.024332721,  7.17928981, 7.159898654, 7.009977785, 6.830979234,
  2.204955099, 2.331625217, 2.133305231,  2.18332885,  2.12778313, 2.269697813, 2.264705552, 2.253940441, 2.287924323,
   7.28437278, 6.983593721,  6.86337111, 6.865970678, 7.219840938, 7.181113053, 7.392230178, 7.484052914,  7.52498281,
  4.265792764, 4.970684112, 4.595545125, 4.575545289, 4.547957809,  4.68215122, 4.674495889, 4.675841709, 4.643311767,
    2.6943516, 2.916324936, 2.578130269, 2.659717988, 2.567436676,   2.8095128, 2.790110381, 2.795882913, 2.884588792,
  3.646303109, 8.817891552,  11.4248793, 10.74738082, 9.296043108,  9.53150669, 8.285160496, 9.769919327, 9.774610531,
  3.040292001,  3.38486713, 2.958851115, 3.047880699, 2.878562717, 3.209319974,  3.20260379, 3.195993624,   3.3004227,
  2.357625231, 2.444753172, 2.340767158,  2.32143889, 2.282608342, 2.401218719, 2.385568421, 2.375334953, 2.432634747,
  5.378494673, 6.065038394, 5.134842087, 5.367342376, 5.682051149, 5.712072512,  5.57179966,  5.72082395, 5.656674512,
  2.833814735, 3.038434511, 2.837711812, 2.859800224, 2.866040813, 2.969167906, 2.929449968, 2.963530689, 2.931065261,
  6.192932281, 6.478439634, 6.180169144, 6.151689376, 6.238949956, 6.708196123, 6.441437631, 6.448280595, 6.413562269,
  4.543042482, 4.786227217, 4.445131477,  4.51471011, 4.491645167, 4.460114204, 4.602482637, 4.587221948, 4.623125028,
  6.069437462, 6.232738284,  6.74644117,  7.04995802, 6.938928532, 6.348253102, 6.080950712, 6.324619355, 6.472893789
  )

Answer 4

一个基本的 R方法。

df_new <- data.frame(t(df1), ID=colnames(df1))
df_new <- merge(df, df_new, by = 'ID')
out <- apply(df_new[, grep('^X', names(df_new))], 1, \(x) {
    data.frame(min=min(x), IQR_low=quantile(x, .25),
          mean=mean(x), median=median(x),IQR_high=quantile(x, .75),
          max=max(x), sd=sd(x))
})
out <- round(do.call(rbind, out), 2L)
rownames(out) <- df_new$ID

Output

> out
           min IQR_low mean median IQR_high   max   sd
GSM239170 2.20    2.97 4.56   4.24     6.10  7.98 1.91
GSM239323 2.33    3.22 5.04   4.98     6.49  8.82 1.98
GSM239324 2.13    2.91 4.93   4.52     6.48 11.42 2.45
GSM239326 2.18    2.91 4.97   4.66     6.57 10.75 2.34
GSM239328 2.13    2.88 4.91   4.69     6.56  9.30 2.15
GSM239329 2.27    3.11 4.95   4.57     6.48  9.53 2.13
GSM239331 2.26    2.97 4.88   4.78     6.43  8.29 1.96
GSM239332 2.25    3.08 4.97   4.63     6.39  9.77 2.16
GSM239333 2.29    3.04 5.01   4.75     6.47  9.77 2.14

数据

df <- structure(list(ID = c("GSM239170", "GSM239323", "GSM239324", 
"GSM239326", "GSM239328", "GSM239329", "GSM239331", "GSM239332", 
"GSM239333"), Disease = c("Control", "Control", "Control", "Control", 
"AML", "AML", "AML", "Control", "Control")), row.names = c(NA, 
-9L), class = "data.frame")

df1 <- structure(list(GSM239170 = c(3.016704177, 7.977735461, 4.207280707, 
7.25609471, 2.204955099, 7.28437278, 4.265792764, 2.6943516, 
3.646303109, 3.040292001, 2.357625231, 5.378494673, 2.833814735, 
6.192932281, 4.543042482, 6.069437462), GSM239323 = c(3.285669072, 
6.532514237, 4.994965767, 7.420807337, 2.331625217, 6.983593721, 
4.970684112, 2.916324936, 8.817891552, 3.38486713, 2.444753172, 
6.065038394, 3.038434511, 6.478439634, 4.786227217, 6.232738284
), GSM239324 = c(2.929482692, 6.388007183, 4.40159671, 6.999340125, 
2.133305231, 6.86337111, 4.595545125, 2.578130269, 11.4248793, 
2.958851115, 2.340767158, 5.134842087, 2.837711812, 6.180169144, 
4.445131477, 6.74644117), GSM239326 = c(2.922820483, 6.466679556, 
4.747114589, 7.094488581, 2.18332885, 6.865970678, 4.575545289, 
2.659717988, 10.74738082, 3.047880699, 2.32143889, 5.367342376, 
2.859800224, 6.151689376, 4.51471011, 7.04995802), GSM239328 = c(3.15950317, 
6.432795021, 4.830045513, 7.024332721, 2.12778313, 7.219840938, 
4.547957809, 2.567436676, 9.296043108, 2.878562717, 2.282608342, 
5.682051149, 2.866040813, 6.238949956, 4.491645167, 6.938928532
), GSM239329 = c(3.163327169, 6.407321524, 4.213762092, 7.17928981, 
2.269697813, 7.181113053, 4.68215122, 2.8095128, 9.53150669, 
3.209319974, 2.401218719, 5.712072512, 2.969167906, 6.708196123, 
4.460114204, 6.348253102), GSM239331 = c(2.985901308, 6.426470803, 
4.884418365, 7.159898654, 2.264705552, 7.392230178, 4.674495889, 
2.790110381, 8.285160496, 3.20260379, 2.385568421, 5.57179966, 
2.929449968, 6.441437631, 4.602482637, 6.080950712), GSM239332 = c(3.122708843, 
6.376394357, 4.4318876, 7.009977785, 2.253940441, 7.484052914, 
4.675841709, 2.795882913, 9.769919327, 3.195993624, 2.375334953, 
5.72082395, 2.963530689, 6.448280595, 4.587221948, 6.324619355
), GSM239333 = c(3.070948463, 6.469070308, 4.849665444, 6.830979234, 
2.287924323, 7.52498281, 4.643311767, 2.884588792, 9.774610531, 
3.3004227, 2.432634747, 5.656674512, 2.931065261, 6.413562269, 
4.623125028, 6.472893789)), row.names = c(NA, -16L), class = "data.frame")